130_notes.dvi

Numbering the three regions from left to right,

u 1 (x) =eikx+Re−ikx

u 2 (x) =Aeik

′x

+Be−ik

′x

u 3 (x) =Teikx

Again we have assumed abeam of definite momentum incident from the left and no wave
incident from the right.

0

x

0

-V 0

Energy

-ik’x

V(x)

incident wave

reflected

E

transmitted

e + Re Te

Ae + Be

ikx -ikx ikx

ik’x

There arefour unknown coefficients. We now match the wave function and its first derivative
at the two boundaries yielding4 equations.

Some hard work yields reflection and transmission amplitudes (See section 9.7.2)

R = ie−^2 ika

(k′^2 −k^2 ) sin(2k′a)

2 kk′cos(2k′a)−i(k′^2 +k^2 ) sin(2k′a)

T = e−^2 ika

2 kk′

2 kk′cos(2k′a)−i(k′^2 +k^2 ) sin(2k′a)

.

The squares of these givethe reflection and transmission probability, since the potential is
the same in the two regions.

Again,classically, everything would be transmittedbecause the energy is larger than the
potential. Quantum mechanically, there is a probability to be transmitted and a probability to be
reflected. Thereflection probability will go to zero for certain energies:R→0 if

2 k′a=nπ